Use properties of basic geometric shapes (triangles, rectangles, circles) - TACHS Math

1. Refer to the figure. A semicircle is attached to one of the 10-meter sides of a 10 m by 4 m rectangle to make a garden bed in the shape shown. Using $\pi \approx 3.14,$ what is the total area, in square meters, of the garden bed?
   Answer: 79.3 m² Rectangle area $=10\times4=40\text{ m}^2.$ The semicircle has radius 5 m, so area $=\tfrac12\pi r^{2}=\tfrac12(3.14)(25)=39.25\text{ m}^2.$ Total $=40+39.25\approx79.3\text{ m}^2.$ B: 65.0 m² leaves out most of the semicircle's area. C: 75.4 m² uses an incorrect radius of 4 m for the semicircle. D: 90.0 m² overestimates by treating the added shape as a full circle instead of a half circle.
2. Refer to the figure. In circle $O,$ $\overline{AB}$ is a diameter and point $C$ lies on the circle. What is the measure of $\angle ACB$?
   Answer: 90° An angle inscribed in a semicircle is a right angle, so $m\angle ACB=90^{\circ}.$ A: 60° might be chosen thinking the triangle is equilateral, but no equal sides are shown. C: 120° misapplies the theorem to give an obtuse angle. D: 180° is the measure of the diameter, not the inscribed angle.
3. In the figure, rectangle $PQRS$ has dimensions $PQ = 20$ and $QR = 15$. Diagonals $\overline{PR}$ and $\overline{QS}$ intersect at point $M$. What is the length of $\overline{QM}$?
   Answer: 12.5 Diagonal length = $\sqrt{20^2+15^2}=\sqrt{625}=25$. In a rectangle, diagonals bisect each other AND are equal, so $QM$ = half of diagonal = 12.5. Distractor A is half of 20. C is half of 15. D is diagonal minus 7.5.
4. Refer to the figure. $\triangle GHI$ has two angles labeled $35^{\circ}$ and $65^{\circ}$ as shown. What is the measure of the unlabeled angle at vertex $H$?
   Answer: 80° Sum of angles in a triangle is $180^{\circ}$. $35^{\circ}+65^{\circ}+x=180^{\circ}\Rightarrow x=80^{\circ}.$ A: 45° assumes a right triangle incorrectly. B: 60° results from subtracting from 125° instead of 180°. D: 90° misinterprets the drawing as right-angled.
5. Refer to the figure. A right triangle has legs of length 9 and 12. A circle is inscribed in the triangle (tangent to all three sides). What is the radius of the inscribed circle?
   Answer: 3 Hypotenuse = $\sqrt{81+144}=15$. For any triangle, $r = \text{Area}/s$ where $s$ is semi-perimeter. Area = $\frac{1}{2}(9)(12)=54$. Semi-perimeter = $(9+12+15)/2 = 18$. So $r = 54/18 = 3$. (Alternatively, for right triangle: $r = (a+b-c)/2 = (9+12-15)/2 = 3$.) Distractor A uses $(a+b-c)/3$. C uses $a+b-c$ divided wrongly. D is $9/2$.
6. Refer to the figure. What is the length, in centimeters, of diagonal $\overline{AC}$ of rectangle $ABCD$?
   Answer: 10 cm Rectangle sides are perpendicular, so $\triangle ABC$ is right with legs 8 cm and 6 cm. Using the Pythagorean Theorem: $AC=\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\sqrt{100}=10\text{ cm}.$ B: 9 cm comes from adding 3 to the shorter side, not using the theorem. C: 11 cm results from adding 3 to the longer side, also incorrect. D: 12 cm is the hypotenuse of a 5–12–13 triangle, but that does not match the given legs.
7. In the figure, triangle $ABC$ is equilateral with side length 12. Point $D$ lies on $\overline{BC}$ such that $BD = 4$. What is the length of $\overline{AD}$?
   Answer: $4\sqrt{7}$ Using the Law of Cosines in triangle $ABD$: $AD^2 = AB^2 + BD^2 - 2(AB)(BD)\cos(60°) = 144 + 16 - 2(12)(4)(0.5) = 160 - 48 = 112$. So $AD = \sqrt{112} = 4\sqrt{7}$. Distractor B results from forgetting the subtraction: $\sqrt{160} = 4\sqrt{10}$. C is $\sqrt{80}$ from using $\cos(60°) = 0$ incorrectly. D assumes the triangle splits into two right triangles incorrectly.
8. In the figure, $\overline{AB}$ is a diameter of the circle with center $O$, and $C$ is a point on the circle. If the measure of arc $BC$ is $70°$, what is the measure of angle $OCA$?
   Answer: $35°$ Arc $BC = 70°$ means central angle $BOC = 70°$. Since $AB$ is a diameter, angle $AOC = 180° - 70° = 110°$. Triangle $AOC$ is isosceles ($OA = OC$ = radius), so the base angles $OAC$ and $OCA$ are equal: each is $(180° - 110°)/2 = 35°$. Distractor B is $90° - 35°$. C is the arc measure. D is the central angle $AOC$.
9. In the figure, $\overline{PA}$ and $\overline{PB}$ are tangent to the circle at points $A$ and $B$, respectively. If $PA = 15$ and the radius of the circle is 8, what is the distance from $P$ to the center $O$ of the circle?
   Answer: 17 The radius $\overline{OA}$ is perpendicular to tangent $\overline{PA}$ at $A$. Triangle $OAP$ is right-angled at $A$: $OP = \sqrt{15^2 + 8^2} = \sqrt{225+64} = \sqrt{289} = 17$. Distractor B uses $15^2-8^2$ incorrectly. C uses the sum $15+8$. D uses an incorrect combination of the given values.
10. Refer to the figure. A regular hexagon has a side length of 6. What is the area of the hexagon?
    Answer: $54\sqrt{3}$ A regular hexagon with side $s$ has area $\frac{3\sqrt{3}}{2}s^2 = \frac{3\sqrt{3}}{2}(36) = 54\sqrt{3}$. Equivalently, 6 equilateral triangles with side 6: $6 \cdot \frac{\sqrt{3}}{4}(36) = 54\sqrt{3}$. Distractor A uses 4 triangles. C uses 2 triangles. D uses 8 triangles.
11. Refer to the figure. Two concentric circles have radii 5 and 13. A chord of the larger circle is tangent to the smaller circle. What is the length of this chord?
    Answer: 24 The radius to the tangent point (length 5) is perpendicular to the chord and bisects it. Using Pythagorean theorem: half-chord = $\sqrt{13^2 - 5^2} = \sqrt{144} = 12$. Full chord = 24. Distractor A is half the chord. B adds radii. D is the diameter of the large circle.
12. Refer to the figure. The shaded region is an annulus (ring) formed by two concentric circles with radii 10 cm and 6 cm. Using $\pi \approx 3.14,$ what is the area, in square centimeters, of the shaded region?
    Answer: 201.0 cm² Outer area $=\pi(10^{2})=3.14(100)=314.$ Inner area $=\pi(6^{2})=3.14(36)=113.0.$ Shaded area $=314-113=201\text{ cm}^2.$ B: 314.0 cm² is the whole outer circle. C: 157.0 cm² averages the two areas instead of subtracting. D: 113.0 cm² is only the inner circle.
13. Refer to the figure. Using $\pi \approx 3.14,$ what is the circumference, in meters, of the circle?
    Answer: 43.96 m The diameter is 14 m, so the radius is 7 m. Circumference $C=2\pi r=2(3.14)(7)=43.96\text{ m}.$ B: 87.92 m doubles the correct result, as though multiplying by 4π. C: 21.98 m uses πr instead of 2πr. D: 28.00 m uses 2r without π.
14. Refer to the figure. The shaded L-shaped patio is made by removing a 5 m by 4 m rectangle from the corner of an 11 m by 9 m rectangle. What is the area, in square meters, of the patio?
    Answer: 79 m² Large rectangle area: $11\times9=99\text{ m}^2.$ Removed rectangle area: $5\times4=20\text{ m}^2.$ Patio area $=99-20=79\text{ m}^2.$ B: 60 m² subtracts 39 instead of 20. C: 99 m² is the full rectangle with nothing removed. D: 20 m² is only the removed piece, not the patio.
15. Refer to the figure. A sector of a circle with radius 9 cm has a central angle of $120^{\circ}$. Using $\pi \approx 3.14,$ what is the area, in square centimeters, of the shaded sector?
    Answer: 84.8 cm² Sector area $=\dfrac{120^{\circ}}{360^{\circ}}\times\pi r^{2}=\dfrac13(3.14)(9^{2})=\dfrac13(3.14)(81)=\dfrac13(254.34)\approx84.8\text{ cm}^2.$ A: 56.5 cm² uses a 90° sector. C: 188.5 cm² uses 240° instead of 120°. D: 254.3 cm² is the full-circle area, not a sector.
16. In the figure, a circle with radius 10 has a chord $\overline{AB}$ of length 16. What is the distance from the center of the circle to the chord?
    Answer: 6 The perpendicular from the center bisects the chord, creating a right triangle with hypotenuse 10 (radius) and one leg 8 (half of chord). Other leg = $\sqrt{100-64}=6$. Distractor A halves the radius. C is half the chord. D uses $\sqrt{100-16}$ (forgetting to halve the chord).

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